You need to show what you have tried, even if it is wrong. That will enable us to help you better.
Note that, for a polynomial with real coefficients, if we have a complex root then we also have the complex conjugate as a root as well. So your cubic polynomial is \(\displaystyle f(x) = a( x - (-1))(x - (2i))(x - (-2i))\).
(The "a" is there because the polynomial will have the same roots no matter what constant is multiplying the whole thing. Your answer key is setting a = 1.)
Edit: The method worked to give the right answer, but the signs were wrong. The terms for the given roots are of the form x - r, not x + r as I had originally written.
Substitute $x=-1$ and $x=2i$ in each of the given expressions. Do you get $0$? If the same expression gives $0$ for both these two values of $x$, then the equation is the one you’re looking for; otherwise, it isn’t.
Right, let’s do it one at a time. Start with $x=-1$. Substitute this in each of the five expressions A–E. You’ll find that B and D give you $0$ whereas A, C and E don’t. Therefore the correct answer is either B or D (you have eliminated A, C and E).
Now substitute $x=2i$ in B and in D. One will give you $0$ while the other won’t. The one that gives you $0$ is the one you’re looking for.
First, there exist an infinite number of different cubic equations that have "-1" and "2i" as roots, Notice that all of the proposed solutions have real coefficients. A polynomial equation with real coefficients but a non-real root must also have the complex conjugate of that non-real root as a root. The complex conjugate of "2i" is "-2i" so the polynomial must be [tex](x+ 1)(x- 2i)(x+ 2i)= (x- 1)(x^2+ 4)= x^3- x^2+ 4x- 4[/tex].